A magnetic field is uniform in a region of radius 10cm, and increasing at 0.02 Tesla/second. What is the magnitude of electric field in the region outside the magnetic field at radius of 15 cm?

To solve this problem, we need to use Faraday's law of electromagnetic induction, which states that the magnitude of the electric field induced by a changing magnetic field is directly proportional to the rate of change of the magnetic field.

Given:
Radius of the uniform magnetic field (r1) = 10 cm = 0.10 m
Rate of increase of magnetic field (dB/dt) = 0.02 Tesla/second
Radius outside the magnetic field (r2) = 15 cm = 0.15 m

Using Faraday's law, we have the equation:

E = -(dΦ/dt) / (2πr)

Where E is the magnitude of the electric field, dΦ/dt is the rate of change of magnetic flux, and r is the distance from the center of the magnetic field.

To find the rate of change of magnetic flux (dΦ/dt), we multiply the rate of change of the magnetic field (dB/dt) by the area of the magnetic field.

The area of the magnetic field can be calculated using the formula:

A = πr^2

Substituting the values:

A = π(0.10 m)^2
A = π(0.01 m^2)
A = 0.0314 m^2

Now, we can calculate the rate of change of magnetic flux:

dΦ/dt = dB/dt * A
dΦ/dt = 0.02 Tesla/second * 0.0314 m^2
dΦ/dt ≈ 0.000628 Weber/second

Finally, we can calculate the magnitude of the electric field outside the magnetic field at radius 15 cm:

E = -(dΦ/dt) / (2πr)
E = -(0.000628 Weber/second) / (2π * 0.15 m)
E ≈ -0.00133 N/C

Therefore, the magnitude of the electric field outside the magnetic field at a radius of 15 cm is approximately 0.00133 N/C. Note that the negative sign indicates that the direction of the electric field is opposite to the direction of the changing magnetic field.

To find the magnitude of the electric field outside the magnetic field, you can use Faraday's law of electromagnetic induction. According to Faraday's law, the magnitude of the electric field induced around a closed loop is equal to the rate of change of magnetic flux through the loop.

In this case, we have a uniform magnetic field that is increasing at a constant rate. Since the magnetic field is uniform, the magnetic flux through a circular loop outside the magnetic field can be calculated by multiplying the magnetic field strength by the area of the loop.

Given:
Radius of the magnetic field region (r1) = 10 cm
Rate of change of the magnetic field (dB/dt) = 0.02 Tesla/second
Radius outside the magnetic field (r2) = 15 cm

First, we need to calculate the magnetic flux at radius r1 and at a specific point in time. The formula for magnetic flux is:

Φ = B * A

Where:
Φ is the magnetic flux
B is the magnetic field strength
A is the area of the loop

At radius r1, the area of the loop is:

A = π * (r1^2 - 0^2) = π * (0.1^2 - 0^2) = π * 0.01 m²

The magnetic flux at radius r1 is:

Φ1 = B1 * A = B1 * π * 0.01 m²

Now, we need to calculate the magnetic flux at radius r2, which is outside the magnetic field. The area of the loop at r2 is:

A = π * (r2^2 - r1^2) = π * (0.15^2 - 0.1^2) = π * 0.025 m²

The rate of change of the magnetic flux at r2 is equal to the rate of change of the magnetic field:

dΦ/dt = dB/dt

Now, we can substitute the values into the equation and solve for the magnitude of the electric field (E):

E = |dΦ/dt| / A

E = |dB/dt| / A

E = |0.02 Tesla/second| / (π * 0.025 m²)

Calculating this, we get:

E ≈ 0.32 Tesla/m²

Therefore, the magnitude of the electric field outside the magnetic field at a radius of 15 cm is approximately 0.32 Tesla/m².